triangles in $\mathbb{R}^n$ with all vertices in $\mathbb{Q}^n$

I have a positive answer if the triangle is not obtuse. Hopefully, the proof idea can be extended.

First of all, by rescaling with some big positive integer, we may assume that $a^2, b^2, c^2$ are all even positive integers. So assume that the triangle is not obtuse, i.e., WLOG we have $a^2\leq b^2\leq c^2\leq a^2+b^2$.

Let $n:= \frac{a^2+b^2+c^2}{2}$ be the dimension. Vertex $C$ is the origin. Vertex $B$ is the point whose FIRST $a^2$ coordinates are $1$ and the rest is $0$. The third vertex $A$ is the point whose LAST $b^2$ coordinates are $1$ and the rest is $0$. Then the length of $BC$ is $a$, the length of $AC$ is $b$, and the length of $AB$ is $\sqrt{n-(a^2+b^2-n)}= \sqrt{2n-(a^2+b^2)}= c$.

The answer is yes, and as quasi pointed out, it is also possible to solve the problem in at most 12 dimensions in the non-obtuse case. I will modify that construction, and provide the full proof in the obtuse case, which is more complicated. So WLOG assume that $a^2\leq b^2\leq c^2$ are all positive even integers.

Case 1: The triangle is not obtuse, i.e., $c^2\leq a^2+b^2$. Let $s:= \frac{a^2+b^2-c^2}{2}\in \mathbb{N}$, and let $x:= b^2-s\in \mathbb{N}, y:= a^2-s\in \mathbb{N}$. By Lagrange's four-square theorem (see https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem ), it is possible to write $x,y,s$ as the sum of squares of 4 integers. Produce vectors of length 4 out of these quartets of numbers, and denote them by $\underline{x}, \underline{y}, \underline{s}$, respectively.

Let $C$ be the origin. Define $A,B$ as the concatenation of vectors of length four as follows:

$$A=(\underline{x},\underline{s},\underline{0})$$ $$B=(\underline{0},\underline{s},\underline{y})$$

Then $\vert AB\vert=c$, $\vert BC\vert=a$, $\vert CA\vert=b$.

Case 2: The triangle is obtuse, i.e., $c^2> a^2+b^2$. We reduce this to the non-obtuse case. We pick positive rational numbers $x,y,z$ such that $c^2-z^2< (a^2-x^2)+(b^2-y^2)$, and $a_1:= \sqrt{a^2-x^2}, b_1:= \sqrt{b^2-y^2}, c_1:=\sqrt{c^2-z^2}$ are sides of a triangle in increasing order, and $x+y=z$. To find such rational numbers, we introduce the variable $\lambda>0$ and solve:

$x=\lambda a$, $y=\lambda b$, $z=x+y$, and $c^2-z^2= (a^2-x^2)+(b^2-y^2)$.

By substituting $z=x+y$ in the last equation we get: $$2xy= c^2-a^2-b^2= -2ab\cos \gamma$$ (Note that this is positive as the triangle is obtuse.) Putting $x=\lambda a$, $y=\lambda b$ yields the solution $\lambda= \sqrt{-\cos \gamma}$, which satisfies $0<\lambda<1$. As rational numbers are dense in the reals, we can pick rational numbers $x\approx \lambda a, y\approx \lambda b, z:= x+y$, such that the equation is ruined in the right direction to obtain $c^2-z^2< (a^2-x^2)+(b^2-y^2)$.

Now we can solve the problem using 13 coordinates. First, produce the solution in 12 coordinates for the triangle $(a_1, b_1, c_1)$. This is possible, as $a_1^2, b_1^2, c_1^2$ are rational and the triangle is acute, so we can refer to Case 1. The solution is $A_1, B_1, C_1$ where $C_1$ is the origin. We extend these solution vectors by a lucky thirteenth coordinate. In case of $C$, the new coordinate is also $0$, so $C$ is the origin in $\mathbb{R}^{13}$. In case of $A$, the new coordinate is $y$, and in case of $B$, the new coordinate is $-x$. Then $AC^2=b_1^2+y^2=b^2$, $BC^2=a_1^2+x^2=a^2$, and $AB^2=c_1^2+(x+y)^2= c_1^2+z^2= c^2$.

Let me also point out that at least 4 dimensions is necessary in general, as there exist rational numbers that cannot be written as the sum of squares of three rational numbers.